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samnatha

  • one year ago

in a particular game a player throws two fair dice together. if their total is 7 or 11 he wins if their total is 2,3 or 12 he looses for any other total, he neither wins or looses. (i) find the probability of winning (ii) find the probability of losing

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  1. tcarroll010
    • one year ago
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    Start with listing all your possible dice pair outcomes and count the number of 7's and 11's out of all possible outcomes. That gives your probability of winning. Same process with the losing for 2, 3, and 12.

  2. tcarroll010
    • one year ago
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    So start with "all possible outcomes". How many of those are there? hint: each die has 6 faces.

  3. samnatha
    • one year ago
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    am 2 dice so 12 ?

  4. tcarroll010
    • one year ago
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    Better start with one die. How many outcomes?

  5. samnatha
    • one year ago
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    6 ?

  6. tcarroll010
    • one year ago
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    Yes. Now for EACH GIVEN outcome on that first die, you can have 6 outcomes on the second die. So, 6 x 6 = 36 possible outcomes for a roll of 2 die. Now, it really would be beneficial for you to write out on paper a 2-dimensional table where horizontal is one die and vertical is the other die and make a table of 36 rolls. That will benefit you immensely here.

  7. samnatha
    • one year ago
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    is the answer for part (i) 2over 9 ?

  8. tcarroll010
    • one year ago
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    Yes, 2/9 for winning. That comes from 6 7's and 2 11's inning rolls out of a possible 36: 8/36 = 2/9 Good job!

  9. samnatha
    • one year ago
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    yeah ok thank you is is the same method for loosing aswell ?

  10. tcarroll010
    • one year ago
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    Very similar. Add up all 2's (there's just 1), all 3's (there are 2) and all 12's (there's 1) for that total over 36.

  11. samnatha
    • one year ago
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    1 over 9 ?

  12. tcarroll010
    • one year ago
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    Yes, because 4/36 = 1/9. This would be a nice game to play because you win more than lose!

  13. samnatha
    • one year ago
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    thank you :) i actuallt have a similar question to that and i was wondering if i got it right or not do you mind checking for me ?

  14. tcarroll010
    • one year ago
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    Good luck in all of your studies and thx for the recognition! @samnatha I'll check it if you like.

  15. samnatha
    • one year ago
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    thank you i think i sent to you in mail :)

  16. samnatha
    • one year ago
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    i'll just put it here aswell :)

  17. samnatha
    • one year ago
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    this circle is divided into 6 equal sectors. you pay €10 to spin the arrow and you win the amount in the sector where the arrow stops. what is the expected amount you win or loose in this game ? theres a picture of a cirlce aswell divided into 6 sections in each section there is different numbers which are €0 €30 €10 €10 €0 and €5

  18. tcarroll010
    • one year ago
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    Since you pay 10 to play, we are really talking about the "net" winning or losing, so you can actually re-do the 6 sectors as a "net" amount: -10, +20, 0, 0, -10, -5 You add those up and divide by 6 to get -5/6. Not a good game to play. Over time, you lose big.

  19. samnatha
    • one year ago
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    ok i'm slighlt confused on how you did all that not great at probability

  20. tcarroll010
    • one year ago
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    np. Each sector is equally likely, and for each sector, we can conceptually see ourselves handing over 10 (at first, to play) and then getting back some money (if any for that sector). So, for example, if we landed on "get 30", it's a net of 20 because we had to pay 10 up front. Your expected value is the net amount times the probability of landing in that sector. It is 1/6 times each sector and then all added up. Or you could just add them all up first: (1/6)(-10) + (1/6)(20) + (1/6)(0) + (1/6)(0) + (1/6)(-10) + (1/6)(-5) or what I did: (1/6)(-10 + 20 + 0 + 0 + -10 -5)

  21. samnatha
    • one year ago
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    ooh ok thank you i am use to doing it the other way so that made more sense thanks alot :)

  22. tcarroll010
    • one year ago
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    yw!

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